本題要求編寫程序，計(jì)算 2 個(gè)有理數(shù)的和、差、積、商。

輸入格式：

輸入在一行中按照?a1/b1 a2/b2?的格式給出兩個(gè)分?jǐn)?shù)形式的有理數(shù)，其中分子和分母全是整型范圍內(nèi)的整數(shù)，負(fù)號(hào)只可能出現(xiàn)在分子前，分母不為 0。

輸出格式：

分別在 4 行中按照?有理數(shù)1 運(yùn)算符 有理數(shù)2 = 結(jié)果?的格式順序輸出 2 個(gè)有理數(shù)的和、差、積、商。注意輸出的每個(gè)有理數(shù)必須是該有理數(shù)的最簡(jiǎn)形式?k a/b，其中?k?是整數(shù)部分，a/b?是最簡(jiǎn)分?jǐn)?shù)部分；若為負(fù)數(shù)，則須加括號(hào)；若除法分母為 0，則輸出?Inf。題目保證正確的輸出中沒有超過整型范圍的整數(shù)。

輸入樣例 1：

2/3 -4/2

輸出樣例 1：

2/3 + (-2) = (-1 1/3) 2/3 - (-2) = 2 2/3 2/3 * (-2) = (-1 1/3) 2/3 / (-2) = (-1/3)

輸入樣例 2：

5/3 0/6

輸出樣例 2：

1 2/3 + 0 = 1 2/3 1 2/3 - 0 = 1 2/3 1 2/3 * 0 = 0 1 2/3 / 0 = Inf 作者: CHEN, Yue 單位: 浙江大學(xué) 時(shí)間限制: 200 ms 內(nèi)存限制: 64 MB 代碼長(zhǎng)度限制: 16 KB

?

#include <iostream> #include <stdio.h> #include <algorithm> #include <string> #include <map> using namespace std; const int maxn = 100010; struct num{long long k = 0;long long up;long long down; }; int gcd(long long a, long long b){return b == 0 ? a : gcd(b, a % b); } num add(num n1, num n2){num res;res.down = n1.down*n2.down;res.up = n1.down*n2.up + n1.up*n2.down;return res; } num sub(num n1, num n2){num res;res.down = n1.down*n2.down;res.up = n2.down*n1.up - n2.up*n1.down;return res; } num mul(num n1, num n2){num res;res.down = n1.down*n2.down;res.up = n1.up*n2.up;return res; } num dive(num n1, num n2){num res;res.down = n1.down*n2.up;res.up = n1.up*n2.down;if (res.down < 0){res.down = -1 * res.down;res.up = -1 * res.up;}return res; } void clean(num n){int flag = 0;n.k = n.up / n.down;if (n.up < 0){n.up = -n.up; flag = 1;}n.up = n.up%n.down;int g = abs(gcd(n.up, n.down));n.up /= g;n.down /= g;if (flag == 1){printf("(");}if (n.k != 0){printf("%lld", n.k);if (n.up != 0){printf(" %lld/%lld", n.up, n.down);}}else{if (n.up != 0){if (flag == 1)printf("-");printf("%lld/%lld", n.up, n.down);}else{printf("0");}}if (flag == 1){printf(")");}} int main(){num n1, n2;scanf("%lld/%lld %lld/%lld", &n1.up, &n1.down, &n2.up, &n2.down);/*int g = abs(gcd(n1.up, n1.down));n1.up /= g;n1.down /= g;g = abs(gcd(n2.up, n2.down));n2.up /= g;n2.down /= g;*/num q, w, e, r;q = add(n1, n2);clean(n1);printf(" + ");clean(n2);printf(" = ");clean(q);printf("\n");w = sub(n1, n2);clean(n1);printf(" - ");clean(n2);printf(" = ");clean(w);printf("\n");r = mul(n1, n2);clean(n1);printf(" * ");clean(n2);printf(" = ");clean(r);printf("\n");if (n2.up != 0){e = dive(n1, n2);clean(n1);printf(" / ");clean(n2);printf(" = ");clean(e);printf("\n");}else{clean(n1);printf(" / ");clean(n2);printf(" = Inf\n");}system("pause"); }

注意點(diǎn)：這道題的坑有兩個(gè)，一個(gè)是int不夠大，兩個(gè)大int乘起來(lái)就爆了，要用long long，long long 的輸入輸出要用lld。第二個(gè)是約分，求最大公約數(shù)，遍歷就超時(shí)了，要用輾轉(zhuǎn)相除法，一定要記住！！！

轉(zhuǎn)載于:https://www.cnblogs.com/tccbj/p/10363287.html

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